from the way you write Banach space and Hilbert space there is a chance that you read German. In that case I would strongly recommend the first chapter of Werner's Funktionalanalysis, where you find a great discussion of many classes of examples of important vector spaces and their basic special features (unfortunately, I don't know of an equivalent in any other language). It is written at about the same level of difficulty as my answer below. (The rest of that book is also great and worthwhile, but the first chapter specifically addresses things pertinent to your question).
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t.b.May 12 '11 at 9:54

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@Theo: You are right! Thank you very much for that hint! I switch constantly between both languages. I found that oftentimes the part that is missing for a proper understanding in one language is to to be found in the other. There are so many texts in German that are not translated and of course the other way round. BTW: The site of Dirk Werner is to be found here: page.mi.fu-berlin.de/werner99 - there are some free texts, e.g. this one about Fixpunktsätze that seems to be very understandable too: page.mi.fu-berlin.de/werner99/skripte/fixpunkte.pdf
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vonjdMay 12 '11 at 10:41

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Ha! Dann hab' ich also richtig geraten! Yes, that's the guy (and either he hasn't shaved or hasn't updated his photograph for quite some time :)). There is the possibility that your university has subscribed to the Springer online access for books and in that case you can access the book for free as well: this is the link. At least you can have a peek. Thanks for the links, btw.
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t.b.May 12 '11 at 10:58

2 Answers
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It seems to me that you've got the things quite right. I restrict attention to real or complex vector spaces. (Edit: But see the update.)

Given a scalar product $\langle \cdot, \cdot\rangle$ we get a norm by setting $\|x\| = \sqrt{\langle x,x \rangle}$ by an application of the Cauchy-Schwarz inequality.

Conversely, a normed vector space structure comes from a pre-Hilbert space structure (as described in 1.) if and only if the norm satisfies the parallelogram law $$2\|x\|^2 + 2\|y\|^2 = \|x-y\|^2 + \|x+y\|^2.$$ See Arturo's answer to this question for an outline of the proof of the non-trivial direction (the trivial direction is just an exercise in expanding the scalar product). Update: I added a detailed solution to that same thread.

Given a norm on a vector space, we get a metric by putting $d(x,y) = \|x - y\|$.

This answers your question about the hierarchy as follows: We have inclusions

where the second row is obtained from the first row by adding the purely metric property of completeness.

Now let me discuss the fact that these inclusions are in fact strict. Let me start with the second row.

The first observation I wish to make is that for $1 \leq p \lt \infty$ and an arbitrary positive integer $n$ the space $\mathbb{R}^{n}$ or $\mathbb{C}^{n}$ with the $\ell^{p}$-norm
$$\Vert x \Vert_{p} = \Vert(x_{1},\ldots,x_{n})\Vert_{p} = \left(\sum_{k=1}^{n} |x_{k}|^{p}\right)^{1/p}$$
is a Banach space. It is a Hilbert space if and only if $p = 2$ (or $n = 1$ where all $\ell^p$-norms coincide). If $n \geq 2$ and $p \neq 2$ the parallelogram law mentioned above fails: Take $x = (1,0,\ldots,0)$ and $y = (0,\ldots,0,1)$, then $\Vert x\Vert_{p} = 1 = \Vert y\Vert_{p}$ while $\Vert x-y\Vert_{p} = 2^{1/p} = \Vert x + y\Vert_p$, so we get
$$4 = 2\Vert x\Vert_{p}^2 + 2\Vert y\Vert_{p}^{2} \neq \Vert x - y \Vert_{p}^2 + \Vert x + y \Vert_{p}^2 = 2 \cdot 2^{2/p}$$
since $p \neq 2$. So we have already plenty of examples of Banach spaces that are not Hilbert spaces.

The space $\mathbb{R}$ with the metric $d(x,y) = \dfrac{|x-y|}{1+|x-y|}$ is a complete metric vector space (an illuminating exercise I urge you to do!), but clearly $d(0,x)$ does not define a norm (as claimed in another "answer"), since $d(0,\lambda x) \neq |\lambda| d(0,x)$ in general.

So far for the rather easy stuff.

Now let me discuss another example. Let $\mathcal{P}([0,1])$ (not quite a standard notation) be the space of polynomial functions $f(x) = a_{n}x^n + \cdots + a_{1} x + a_0$ of arbitrary degree on the interval $[0,1]$. The $\ell^p$-norm can be defined as above: $\Vert f\Vert_{p} = \left(\sum_{k=1}^{n} |a_{k}|^{p}\right)^{1/p}$.
And taking the polynomials $x$ and $x^2$, we see again that the parallelogram law fails if $p \neq 2$, so again this gives an example of a normed vector space that is not a Hilbert space.

The point of this example is that $\mathcal{P}([0,1])$ is never complete with respect to the $\ell^{p}$-norm.

There are many ways to see this. To keep this post as simple as possible, let me do this completely explicitly: Check that $f_{n}(x) = \sum_{k = 0}^{n} \frac{(-1)^{k}}{(2k+1)!} x^{2k+1}$ (the $(2n+1)$-st Taylor polynomial of $\sin{x}$) forms an $\ell^{p}$-Cauchy sequence and doesn't converge in $\mathcal{P}([0,1])$. A slightly more high-powered approach would be to appeal to the Weierstrass approximation theorem (this requires some trickery), or to the general fact that a Banach space has either finite or uncountable dimension—the latter is a standard consequence of the Baire category theorem (which amounts to even worse trickery in my opinion).

Therefore we have now an example of a normed vector space that isn't complete and for $p = 2$ this is an example of a pre-Hilbert space that isn't a Hilbert space. This family of examples settles strict inclusion $\{\text{Hilbert spaces}\} \subsetneqq \{\text{pre-Hilbert spaces}\}$, $\{\text{Banach spaces}\} \subsetneqq \{\text{normed spaces}\}$ and also $\{\text{pre-Hilbert spaces}\} \subsetneqq \{\text{normed vector spaces}\}$.

There are two inclusions I left open so far and I only gave a somewhat unsatisfactory example of a (complete) metric vector space that isn't a Banach space. Since this is a bit more subtle, I simply make some concluding remarks.

The first observation I would like to mention is that a normed vector space has an abundance of convex open sets. Indeed, the ball of radius $r$ around any point is convex. In a metric vector space, however, there is no reason that there be any convex open sets except the empty set and the space itself—this is due to the fact that the metric is not required to be homogeneous, only translation-invariant. One standard example for this is the space $L_{0}([0,1])$ of all (classes of) Lebesgue measurable functions modulo null sets, equipped with the topology of convergence in measure to be explicit, my preferred metric is $\displaystyle d(f,g) = \int \frac{|f - g|}{1 + |f-g|}$.

The difference between the first row and the second row is actually not that big and usually not of much importance, since there is the process of completion. Indeed, the completion of a metric vector space is again a metric vector space (and complete by definition), the completion of a normed space again has a norm and is thus a Banach space by definition and the completion of a pre-Hilbert space is a Hilbert space (since the parallelogram law holds true in the completion). To find examples that belong to the first row but not to the second, we can simply choose a dense subspace of our favourite (infinite-dimensional) example of the second row.

Finally, let me stress that all the topics above are encompassed in the theory of topological vector spaces of which the theory of locally convex spaces is the richest and most reasonable for most applications that need to go beyond normed vector spaces (e.g. the theory of distributions). But I would say that it is most reasonable to stick to Banach spaces and normed vector spaces first, to which a quite good introduction is given in Rudin's Real and complex analysis or any introduction to functional analysis. A decent knowledge of real analysis (in particular measure theory) seems to be the basic prerequisite for appreciating such a book, however.

Update: Here are two comments to a now deleted answer that deserve to be publicly viewable (I omitted references to the people involved and slightly changed the formatting):

Matt E writes:

[...] the notion of Hilbert space depends on $F$ being either $\mathbb{R}$ or $\mathbb{C}$. It requires a positive definite inner product (not an arbitrary one), and positive definite makes sense only for over $\mathbb{R}$ or $\mathbb{C}$. (And note that in the case of $\mathbb{C}$, it requires that the pairing be Hermitian, i.e. conjugate linear in the second variable, a concept which is particular to $\mathbb{C}$.) Again, in your formula for the norm, the square-root only makes sense because you are taking the square root of a positive real number — in a general field an element may not be a square, and even if it is, there won't be any way to extract a canonical square root. [...]

Usually, there's no distinction made between a metric on a vector space and a norm on one, since we can define a norm by $\|v\| = \rho(v,0)$ or a metric by $\rho(v,w) = \|v-w\|$. A norm is really just a better kind of metric, because it doesn't require one to specify two points- since there's an additive structure on the vector space, there is an obvious origin (i.e., the zero vector), and we can define distances from there. Essentially, a norm is a metric that also defines ``size'' in a non-arbitrary way, and, if you have a metric space, defining a vector space structure on it lets you define a norm in term of the metric.

An inner-product space (pre-Hilbert space) is a vector space with a very special type of norm- one that results from an inner product $\langle \cdot, \cdot \rangle$. The inner product gives a notion of angle, and it also defines a norm by $\|v\| = \sqrt{\langle v, v\rangle}$.

Requiring completion on all the spaces just makes them nicer to work with, since you can assume that any Cauchy sequences converge to something, but there's not a big conceptual leap other than that.

The heirarchy you're looking for is pretty close to your chart- normed spaces are a special type of metric space, and pre-Hilbert spaces are a special type of nomred

The first sentence (and most of the fist paragraph) is completely wrong. There are many metric vector spaces that don't admit a norm at all, e.g. $C^{\infty}(0,1)$, the Schwartz space, etc.
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t.b.May 11 '11 at 13:54

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@Theo: This one is interesting: Could you give an intuitive example of a metric where you don't have a norm? Perhaps I am not thinking abstractly enough but if you can define the distance of a vector from the origin you must also be able to measure the length between two points defined by vectors, or not?!?
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vonjdMay 11 '11 at 14:13

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@vonjd: See here. Theo has already given examples. The wikipedia articles contains further ones.
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RasmusMay 11 '11 at 14:21

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@Calvin: The "norm" you defined from your metric will not be homogeneous in general.
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RasmusMay 11 '11 at 14:25

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Calvin: I trivially edited your answer to remove my downvote since I think that -5 is a bit too harsh (such misconceptions happen to all of us). In my answer I tried to clarify some things.
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t.b.May 12 '11 at 9:58